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Tambaya 2 Rahoto
When the r.m.s value of a source of electricity supply is given as 240v, it means that the peak value of the supply is
Bayanin Amsa
When the r.m.s value of a source of electricity supply is given as 240V, it means that the peak value of the supply is about 340V. The root-mean-square (r.m.s) value of an AC voltage or current is a way to express the effective or average value of that signal. In the case of a sine wave, the r.m.s value is approximately 0.707 times the peak value of the wave. So, to calculate the peak voltage of a 240V r.m.s supply, we can divide the r.m.s value by 0.707. Using this formula, the peak value of a 240V r.m.s supply can be calculated as: Peak value = r.m.s value / 0.707 = 240V / 0.707 ≈ 339.6V Therefore, the peak value of the supply is approximately 340V. It is important to note that this formula assumes a perfect sine wave, and in reality, the peak value of a voltage signal can fluctuate depending on various factors.
Tambaya 3 Rahoto
I. Refraction II. Interference III. Diffraction Which of the above properties are common to all waves?
Bayanin Amsa
The property that is common to all waves is "I. Refraction". Refraction is the bending of a wave as it passes from one medium to another, and it occurs for all types of waves, including light, sound, and water waves. On the other hand, "II. Interference" and "III. Diffraction" are not common to all waves. Interference is the phenomenon of two waves overlapping and producing a resultant wave with a new amplitude and phase, which occurs only for waves that are capable of superposition, such as light and sound waves. Diffraction is the bending of waves around obstacles or through small openings, which also occurs only for waves with wavelengths comparable to the size of the obstacle or opening, such as light and water waves, but not for sound waves with longer wavelengths.
Tambaya 4 Rahoto
I. Refraction II. Interference III. Diffraction
Which of the above properties are common to all waves?
Bayanin Amsa
The properties of refraction, interference, and diffraction are all common to all types of waves. Refraction is the bending of waves as they pass through a medium with a different density. All types of waves, including sound waves, water waves, and light waves, can refract when they encounter a medium with a different density. Interference occurs when two or more waves interact with each other. This can result in constructive interference, where the waves combine to form a larger wave, or destructive interference, where the waves cancel each other out. All types of waves can interfere with each other. Diffraction is the bending of waves around obstacles or through openings. This phenomenon can be observed when sound waves bend around a corner, or when light waves diffract through a small opening, creating a pattern of light and dark bands. All types of waves can diffract. Therefore, the correct answer is "I, II and III," as all three properties are common to all waves.
Tambaya 5 Rahoto
The instrument used to measure the pressure of a gas is
Bayanin Amsa
Barometer is used to measure atmosphere pressure
Thermometer is used to measure temperature
Hygrometer is used to measure relative density of a liquid
Manometer is used to measure the pressure of a gas
Tambaya 6 Rahoto
Calculate the specific latent heat of vaporization of steam of 1.13 x 106J pf heat energy is required to convert 15kg of it to water.
Bayanin Amsa
The specific latent heat of vaporization of steam is the amount of heat energy required to convert one kilogram of water into steam at a constant temperature. The given problem states that 1.13 x 106 J of heat energy is required to convert 15 kg of steam to water. We can use the formula: heat energy = mass x specific latent heat where the mass is 15 kg and the heat energy is 1.13 x 106 J. Rearranging the formula, we get: specific latent heat = heat energy / mass Substituting the values, we get: specific latent heat = 1.13 x 106 J / 15 kg = 7.53 x 104 J/kg Therefore, the specific latent heat of vaporization of steam is 7.53 x 104 J/kg. The correct option is (3): 7.53 x 104 J/kg-1.
Tambaya 7 Rahoto
It is known that a neutron exists in a light atomic nucleus
Which of the following also exists in the nucleus?
Bayanin Amsa
A neutron is a subatomic particle that has no electric charge and is found in the nucleus of an atom along with protons. Therefore, the correct option is: - proton An electron is a negatively charged subatomic particle that orbits around the nucleus of an atom. A β- particle, also known as a beta particle, is an electron or a positron emitted during radioactive decay. Both of these particles are not found in the nucleus of an atom. On the other hand, a proton is a positively charged subatomic particle that is also found in the nucleus of an atom along with neutrons. The number of protons in the nucleus determines the atomic number of the element. Hence, a proton also exists in the nucleus along with a neutron. In summary, the only subatomic particle that exists in the nucleus along with a neutron is a proton.
Tambaya 8 Rahoto
Which of the following is correct about an electric cell?
Bayanin Amsa
About electric cell
Primary cells are used up gradually often the cells are use and it can't be recharged
Secondary cells or accumulated can be recharged and used for a long period of time and their main advantage is that they have low internal resistance
Tambaya 9 Rahoto
The action of blotting paper on ink is due to
Bayanin Amsa
The action of blotting paper on ink is due to capillarity. Capillarity is the ability of a liquid to flow in narrow spaces, such as the fibers of a piece of blotting paper, against the force of gravity. This is made possible because of the combination of the liquid's surface tension, which acts like a skin on the surface of the liquid, and the wicking action of the fibers in the paper. The fibers in the blotting paper wick the ink from the surface it is on, and the ink is held in the fibers due to the surface tension of the liquid. This is why blotting paper is effective in removing excess ink from surfaces, because it pulls the ink into its fibers and holds it there.
Tambaya 10 Rahoto
The instrument to measure the relative density of light is
Bayanin Amsa
The instrument used to measure the relative density of light is a hydrometer. A hydrometer is a device that is designed to measure the density of a liquid by comparing it to the density of water. In order to measure the relative density of light, the hydrometer is placed in a liquid with a known density, such as water. The hydrometer will float at a certain level in the liquid, which indicates the density of the liquid. When measuring the relative density of light, the hydrometer is placed in a liquid that has been made less dense by the addition of a solute, such as sugar or salt. The hydrometer will float at a higher level in the less dense liquid, indicating that the relative density of the light is lower than that of the liquid without the solute. In summary, a hydrometer is the instrument used to measure the relative density of light by comparing it to the density of a liquid with a known density, such as water.
Tambaya 11 Rahoto
A radio station broadcast at a frequency of 600KHZ. If the speed of light in air is 3 × 108ms−1 . Calculate the wavelength of the radio wave
Bayanin Amsa
The formula to calculate the wavelength of a wave is: wavelength = speed of light / frequency where wavelength is the distance between two consecutive peaks or troughs of a wave, frequency is the number of cycles of the wave per second, and the speed of light in air is approximately 3 × 10^8 m/s. In this question, the frequency of the radio wave is given as 600 kHz, which means that it has 600,000 cycles per second. Substituting these values into the formula, we get: wavelength = 3 × 10^8 m/s / 600,000 s^-1 wavelength = 500 m Therefore, the wavelength of the radio wave is 500 meters. The answer is: 5.0 × 10^2 m.
Tambaya 12 Rahoto
Calculate the time taken for a mango fruit that fall from the tree 20 m to the ground [g = 10 ms-2]
Bayanin Amsa
To calculate the time taken for a mango fruit to fall from a tree to the ground, we can use the following formula: distance = (1/2) x acceleration x time^2 Where: distance = 20m (the height of the tree) acceleration = 10 m/s^2 (acceleration due to gravity) time = unknown (what we are solving for) Rearranging the formula, we get: time = square root(2 x distance / acceleration) Plugging in the values we get: time = square root(2 x 20 / 10) time = square root(4) time = 2 seconds Therefore, the time taken for a mango fruit to fall from a tree 20 m to the ground is 2 seconds.
Tambaya 13 Rahoto
Which of the following statements is NOT correct?
Bayanin Amsa
The correct answer is "Molecules of a liquid are stationary" is not correct. This statement is incorrect because the molecules of a liquid are not stationary, they are in constant motion. The molecules in a liquid are able to move past each other, which is why liquids can take the shape of their container. The other statements are all correct. Matter is indeed made up of molecules, and the constant motion of these molecules is what gives matter its physical properties. Brownian motion, which is the random movement of particles in a fluid, is a visible example of the motion of particles in matter and provides evidence for the particle nature of matter.
Tambaya 14 Rahoto
Calculate the light of the image formed by a pinhole camera of length 12cm used to photograph an object 60cm away from the hole and 70cm high
Bayanin Amsa
Magnification = Length of cameradistance of object from pin hole
m = 1260
= 0.2
Also magnification m = Height of imageHeight of object
∴ Height of image = m × height of object
= 0.2 × 70
= 14cm
Tambaya 15 Rahoto
What is the cost of running seven 40 W lamps and five 80 W la mps for 12 hours of the electrical energy cost N7.00kWk?
Bayanin Amsa
To calculate the cost of running the lamps, we need to determine the total energy consumed in kilowatt-hours (kWh) and multiply it by the cost of electrical energy per kWh. First, let's calculate the total power consumption of the lamps: Total power consumption = (7 x 40 W) + (5 x 80 W) = 280 W + 400 W = 680 W Next, let's convert the power consumption to kWh: Energy consumption = (680 W / 1000) x 12 hours = 8.16 kWh Finally, let's calculate the cost of running the lamps: Cost = Energy consumption x Cost of electrical energy per kWh Cost = 8.16 kWh x N7.00/kWh = N57.12 Therefore, the cost of running seven 40 W lamps and five 80 W lamps for 12 hours at an electrical energy cost of N7.00/kWh is N57.12. The correct answer is option D.
Tambaya 16 Rahoto
Which of the following is not a mechanical wave?
Bayanin Amsa
Radio waves are not a mechanical wave. Mechanical waves are waves that require a medium to propagate, meaning they can only travel through a physical material. Examples of mechanical waves include sound waves, water waves, and waves in a closed pipe. Radio waves, on the other hand, are electromagnetic waves and do not require a medium to propagate. They can travel through a vacuum, such as space, and are used for communication technologies like radio and television broadcasting, cell phone signals, and GPS.
Tambaya 17 Rahoto
Which of the following representations is correct for an atom X with 28 electrons and 30 neutrons?
Bayanin Amsa
An atom can be represented thus,
ZAX where Z = mass number
A = atomic number
A = number of proton = number of electrons in a free state = 28
Z = number of protons + number of neutrons = 28 + 30 = 58
5828X is the representation
Tambaya 18 Rahoto
When two mirrors are placed at an angle of 900 to each other, how many images will be formed when an object is placed in front of the mirrors
Bayanin Amsa
When two mirror are inclined at an angle θ, the number of images formed, n = 360θ − 1
∴ since θ = 90o
∴ 360θ − 1
= 4 − 1
= 3
Tambaya 19 Rahoto
Which of the following is not an electromagnetic radiation?
Bayanin Amsa
The answer is: sound waves is not an electromagnetic radiation. Electromagnetic radiation is a type of energy that travels through space at the speed of light in the form of waves. It is made up of oscillating electric and magnetic fields, and it includes a wide range of phenomena such as radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, x-rays, and gamma rays. Sound waves, on the other hand, are not electromagnetic radiation. They are mechanical waves that require a medium to travel through, such as air, water, or solids. When an object vibrates, it creates sound waves that travel through the medium and can be heard by our ears. So, while x-rays, radio waves, and sunlight are all examples of electromagnetic radiation, sound waves are not.
Tambaya 20 Rahoto
Which of the following is a semiconductor
Bayanin Amsa
The semiconductor among the given options is "silicon." Semiconductors are materials that have electrical conductivity between that of a conductor and an insulator. They are widely used in electronics, such as in the construction of computer chips. Copper is a good conductor of electricity and is commonly used in electrical wiring. It is not a semiconductor. Silicon, on the other hand, is one of the most commonly used semiconductor materials in the electronics industry. It has four valence electrons, which allow it to form a crystal lattice structure with other silicon atoms. This structure gives silicon its unique electrical properties that make it useful in electronic devices such as transistors and solar cells. Plastic and iron are not semiconductors. Plastic is an insulator, which means it does not conduct electricity at all, while iron is a conductor, like copper, and does not exhibit the unique electrical properties of a semiconductor.
Tambaya 21 Rahoto
Under which of the following conditions is work done?
Bayanin Amsa
Work Is only done when an object is moved to a distance.
The ans is C because the man pushed the table to a distance. The rest no work is done.
Ans C
Tambaya 22 Rahoto
A constant force of 5N acts for 5 seconds on a mass of 5kg initially at rest. Calculate the final momentum
Bayanin Amsa
Impulse = momentum
Impulse = Force (f) × time (t)
Momentum = mass (m) × velocity (v)
? Ft = Final momentum - initial momentum
FT = mv - mu
Since it is initially at rest u = 0
? 5 × 5 = mv ? m(0)
25 = mv
? Final momentum = 25kgms?1
Tambaya 23 Rahoto
How can energy loss be minimized through Eddy-current?
Bayanin Amsa
Eddy currents are produced by the varying flux cathode, the iron core of an equipment thus reducing efficiency due to power consumption. It can be reduced by laminating the core by breaking up the path of eddy current or by increasing the resistance of the core
Usage of insulated soft iron wire is to reduce hysteresis loss
Usage of low resistance wire (thick wire) is to reduce I2R loss
Usage of thick wire is to reduce leakage heat loss due to leakage of magnetic flux
Thus the correct answer is to use high resistance wire or thin wire
Tambaya 24 Rahoto
Calculate the angle of minimum deviation of a 60o prism of a refractive index [sin-10.75 = 49o]
Bayanin Amsa
For a refractive index () = sin12(A+D)sin12A
D = angle of minimum deviation
A = refractive angle of the prism
1.5 = sin12(60+D)sin12×60
1.5 = sin12(60+D)sin30
Sin 12 (60 + D) = 1.5 * Sin 30
Sin 12 (60 + D) = 0.75
12 (60 + D) = Sin-1 (0.75)
but Sin-1 (0.75) = 49o
12 (60 + D) = 49
60 + D = 2 * 49 = 98o
D = 98o - 60o
D = 38o
Tambaya 25 Rahoto
Calculate the upthrust on an object of volume 50cm3 which is immersed in liquid of density 103kgm-3 [g = 10ms-2]
Bayanin Amsa
To calculate the upthrust on an object immersed in a fluid, we need to use Archimedes' principle which states that the upthrust (buoyant force) experienced by an object partially or fully submerged in a fluid is equal to the weight of the fluid displaced by the object. First, let's convert the volume of the object into meters cubed: 50 cm^3 = (50/100)^3 m^3 = 0.0005 m^3 The weight of the liquid displaced by the object is: density of liquid x volume of liquid displaced x acceleration due to gravity = 103 kg/m^3 x 0.0005 m^3 x 10 m/s^2 = 0.515 N Therefore, the upthrust (buoyant force) acting on the object is 0.515 N. So, the correct answer is option C: 0.5N.
Tambaya 26 Rahoto
Determine the focal length of a thin converging lens if the power is 5.0 dioptres
Bayanin Amsa
The formula for calculating the focal length (f) of a thin converging lens in meters using its power (P) in dioptres is: f = 1/P In this problem, the power of the lens is given as 5.0 dioptres. So, substituting this value in the above formula, we get: f = 1/5.0 = 0.2 meters Therefore, the focal length of the lens is 0.2 meters (or 20 centimeters). This means that when parallel light rays pass through this lens, they will converge to a point that is 0.2 meters (20 centimeters) away from the lens. The smaller the focal length of a lens, the greater its ability to converge light, and the stronger its power. So, the correct answer is 0.2 m.
Tambaya 27 Rahoto
Which is the incorrect formula for a body accelerating uniformly?
Bayanin Amsa
s = 12 ut + at2 is a wrong equation, it should be s = ut + 12 at2
Tambaya 28 Rahoto
Which of the following is a scalar quantity?
Bayanin Amsa
Scalar quantity is a quantity with only magnitude, it has no direction
Tension, weight, impulse are vector quantities because they have direction. Mass is the only scalar quantity there
Tambaya 29 Rahoto
If silicon is doped with phosphorus, what type of semiconductor material will be formed?
Bayanin Amsa
When silicon is doped with a trivalent atom. A p-type semi-conductor will be formed when silicon is doped with a pentavalent atom such as arsenic, a n-type semi-conductor will be formed
Phosphorous is a pentavalent element, so it will produce a o -type semi-conductor
Tambaya 30 Rahoto
Three 3ω resistance are connected in parallel what is the equivalent resistance?
Bayanin Amsa
Resistor connected in a parallel have an equivalence given by
1Reff
= 13
+ 13
+ 13
= 1+1+13 = 33 = 1
Reff = 1ω
Tambaya 31 Rahoto
Calculate the electric field intensity between two plates of potential difference 6.5V when separated by a distance of 35cm.
Bayanin Amsa
The electric field intensity E between two parallel plates of potential difference V separated by a distance d is given by E = V/d. Substituting the given values, we have E = 6.5V/0.35m = 18.57 NC^−1. Therefore, the electric field intensity between the two plates is 18.57 NC^−1. Answer: 18.57NC^−1
Tambaya 32 Rahoto
The net capacitance in the circuit above is
Bayanin Amsa
For capacitance i n parallel, 2µF and 2µF are in parallel,
their equivalence is 2µF and 2µF = 4µF
The 4µF generated is now in series with the remaining 4µF.
The net capacitance for series connection is
1C = 14 + 14 = 1 + 14 = 24
C = 42
= 2µF
Tambaya 33 Rahoto
An element and its isotope only differ in the number of
Bayanin Amsa
An element and its isotope only differ in the number of neutrons. Isotopes are different forms of the same element that have the same number of protons in their nucleus, but a different number of neutrons. The number of protons determines the identity of the element and is called the atomic number, while the number of neutrons in the nucleus is called the mass number. Therefore, the difference between an element and its isotope is only the number of neutrons, not the number of protons or electrons. The number of protons and electrons are the same for all isotopes of a given element.
Tambaya 34 Rahoto
From the diagram above, calculate the energy stored in the capacitor
Bayanin Amsa
c = 8µF, v = 10v
Energy stored = 12 cv2
= 12 × (8 × 10− 6) × 102
= 4 × 10−4
Tambaya 35 Rahoto
A block and tackle pulley arrangement has 6 pulleys in which an effort of 50N supports a load of 200N, calculate the efficiency of the machine
Bayanin Amsa
To calculate the efficiency of the block and tackle pulley arrangement, we need to use the formula: efficiency = (load × distance moved by load) / (effort × distance moved by effort) In this case, the load is 200N and the effort is 50N. Since the load is being lifted, its distance moved is equal to the height it is lifted. Let's assume the load is lifted by a distance of 1 meter. Since the effort is being applied over a greater distance due to the pulley system, let's assume it moves a distance of 6 meters (since there are 6 pulleys). Using these values in the formula, we get: efficiency = (200N × 1m) / (50N × 6m) = 0.6667 or 66.67% Therefore, the efficiency of the block and tackle pulley arrangement is 66.67%. This means that only 66.67% of the effort is used to lift the load, with the rest being lost to friction and other factors.
Tambaya 36 Rahoto
When two objects P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity pf P to Q is
Bayanin Amsa
H = mcθ
HP = HQ, θP = 2θ,
mP = 12 mQ
∴ HP = mQCPθP, also HQ = mQCQθQ
Since HP = HQ
∴ mPcPθP = mQcPθq
Putting in the conditions
12 mQ × cP × 2θQ = mQ × θQ × cQ
∴ cP = cQ
CpCQ = 11
= 1:1
Tambaya 37 Rahoto
A string of length 5cm is extended by 0.04m when a load of 0.8kg is suspended at the end. How far will it extend if a force of 16N is applied? [g = 10ms−2 ]
Bayanin Amsa
The extension of a string when a load is suspended from it is given by Hooke's Law which states that the extension is proportional to the force applied. The proportionality constant is called the spring constant, represented by k. Mathematically, we can express this relationship as: F = kx where F is the force applied, x is the extension, and k is the spring constant. To solve this problem, we need to find the spring constant, k, of the string. We can use the information given in the problem to calculate k as follows: k = F/x where F is the force applied and x is the extension. We are given that the string of length 5cm (which is 0.05m) is extended by 0.04m when a load of 0.8kg is suspended at the end. We can convert the mass to force using the formula F = mg, where g is the acceleration due to gravity (g = 10ms^-2). F = 0.8kg × 10ms^-2 = 8N Using the formula for the spring constant, we get: k = F/x = 8N / 0.04m = 200N/m Now, we can use this spring constant to find the extension when a force of 16N is applied. Again, we can use the formula F = kx and rearrange it to solve for x: x = F/k Plugging in the values, we get: x = 16N / 200N/m = 0.08m Therefore, the string will extend by 0.08m when a force of 16N is applied. The answer is (D) 0.08m.
Tambaya 38 Rahoto
A block and tackle pulley arrangement has 6 pulleys in which an effort of 50N supports a load of 200N, calculate the efficiency of the machine
Bayanin Amsa
Tambaya 39 Rahoto
Under what conditions are cathode rays produced in a discharge tube?
Bayanin Amsa
Cathode rays are streams of electrons that are emitted from the cathode of a discharge tube. A discharge tube is a partially evacuated glass tube with two metal electrodes: a cathode (negative electrode) and an anode (positive electrode). When a voltage is applied across the electrodes, a current flows through the gas in the tube, and cathode rays can be produced. The conditions required to produce cathode rays in a discharge tube are low pressure and high voltage. When the gas inside the tube is at a low pressure, there are fewer gas molecules to collide with the electrons emitted from the cathode. This means that the electrons can travel a longer distance without colliding with other particles, which allows them to form a visible beam of cathode rays. The high voltage applied to the electrodes causes a potential difference between the cathode and anode. This potential difference accelerates the electrons emitted from the cathode towards the anode. The faster the electrons travel, the more kinetic energy they have, and the more likely they are to collide with other gas molecules in the tube. These collisions can cause the gas molecules to become ionized, which can create a cascade effect, leading to the production of more electrons and the formation of a bright beam of cathode rays. Therefore, the correct answer is: Low pressure and high voltage.
Tambaya 40 Rahoto
Which of the following operations can be used to convert an alternative current dynamo into a direct current dynamo
Bayanin Amsa
A d.c generator is one in which its current is allowed to flow in one direction even through it may vary in value an a.c generator can only be made to produce a d.c by replacing the two slip rings with a single split ring or commutator.
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